Ternary scale theorems: Difference between revisions

# If ''n'' is odd, ''s'' = ''a'''''X''' ''b'''''Y''' ''b'''''Z''' is obtained from some mode of the (primitive) MOS ''a'''''X''' 2''b'''''W''' by replacing all the W's successively with alternating '''Y''''s and '''Z''''s (or alternating '''Z''''s and '''Y''''s for the other chirality, fixing the mode of ''a'''''X''' 2''b'''''W'''). The two alternants differ by replacing one '''Y''' with a '''Z'''.

# If ''n'' is odd, ''s'' is pairwise-MOS. That is, the following operations each result in a [[MOS]]: setting '''L''' = '''M''', setting '''L''' = '''s''', and setting '''M''' = '''s'''.

Since ''s'' is generator-offset it is well-formed with respect to the aggregate generator '''g''' = ('''g'''₂ + '''g'''₁). Since '''g'''₁ and '''g'''₂ subtend the same number of steps by the SGA assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those dyads that are "offset" by '''g'''₁ must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence ('''g'''₃ + '''g'''₁), the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''₂ and '''g'''₃ subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''₂ with '''g'''₃ in the next part.

Since ''n'' > 0, these are all distinct by ℤ-linear independence; hence there are at least 4 sizes for ''k''-steps. A 1-step must be reached by stacking an odd number of generators, thus by applying this argument to 1-steps, we see that there must be at least 4 step sizes in some tuning, a contradiction. Thus '''g'''₁ and '''g'''₂ must themselves be step sizes. Thus we see that an even-length SGA ternary scale must be of the form (xy)^''r''xz. (Note that (xy)^''r''xz is not SV3, since it has only two kinds of 2-steps, '''xy''' and '''xz'''.) This proves (1).

In case 2, let ''n'' ≥ 3 and let (2, 1) − (1, 1) = '''g'''₁, (1, 2) − (2, 1) = '''g'''₂ be the two alternants. Let '''g'''₃ be the closing generator after stacking alternating '''g'''₁ and '''g'''₂. Then the generator circle is ('''g'''₁ '''g'''₂)^{floor(''n''/2)} '''g'''₃. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-''k'' subwords of the (linear) word ('''g'''₁ '''g'''₂)^{floor(''n''/2)}, we see that the first two sizes must both occur (''n'' − ''k'')/2 times. This proves (2).

Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has ''j'' + 1 '''W''''s and the perfect generator has ''j'' '''W''''s. Consider the two alternants, detemperings of the generator ''i'''''X''' + ''j'''''W''' of the ''a'''''X''' 2''b'''''W''' MOS ''T''('''X''', '''W''') = ''s''('''X''', '''W''', '''W''') where gcd(''j'', 2''k'') = 1.

and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two MOSes. These correspond to two of the temperings '''X''' = '''Y''', '''Y''' = '''Z''' and '''X''' = '''Z'''. The third tempering follows by symmetry (by taking the other chirality). {{qed}}